The integral of each individual monomial will be integral. First we have the identity
$$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$
so my claim is that
$$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz \in \mathbb Z.$$
Since $\int_0^1 (k+nz)^{n-2i-1}dz=\frac{(k+n)^{n-2i}-k^{n-2i}}{n-2i}$ and $n$ divides $(k+n)^{n-2i}-k^{n-2i}$, it suffices to show that
$$\frac{n}{n-2i}\binom{n-i-1}{i}\in \mathbb Z.$$
However $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ is clearly an integer, and the claim follows.